The equation of a circle $C$ is $x^2+y^2-12x+18y+113 = 0$. What is its center $(h, k)$ and its radius $r$ ?
Solution: To find the equation in standard form, complete the square. $(x^2-12x) + (y^2+18y) = -113$ $(x^2-12x+36) + (y^2+18y+81) = -113 + 36 + 81$ $(x-6)^{2} + (y+9)^{2} = 4 = 2^2$ Thus, $(h, k) = (6, -9)$ and $r = 2$.